Let $a(x)=4x^5-x^3+7$, and $b(x)=2x^5+x^2$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{4x^5-x^3+7}{2x^5+x^2}$ : We divide ${2x^5}$ into ${4x^5}$ to get ${2}$ : $ \hphantom{1567|1444} {2}\\ {{{2x^5}+x^2}}|\overline{{4x^5}-x^3\ +\ 0x^2+7}\\ \hphantom{37........|}\llap{-}\underline{(4x^5+0x^3+2x^2)}\\ \hphantom{37|3....999......}{-x^3-2x^2+7 }\\ $ [What did we do here?] The process stops here because $2x^5+x^2$ is a polynomial of the fifth degree and $-x^3-2x^2+7$ is a polynomial of the third degree. So it follows that ${r(x)}={-x^3-2x^2+7}$, ${q(x)}={2}$, and $ \dfrac{4x^5-x^3+7}{2x^5+x^2}={2}+\dfrac{{-x^3-2x^2+7}}{2x^5+x^2}$ To conclude, $q(x)=2$ $r(x)=-x^3-2x^2+7$